Answer:
0.0928J
Explanation:
the pulling force of spring F=-kx
where x is the displacement from equilibrium position.
energy stored:
[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]
*** Its fine if you know nothing about calculus. Just apply the equation
[tex]U=\frac{kx^{2} }{2}[/tex]
where U is the potential energy of the spring***
put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)
g is incident on 3 successive sheets of polarizing material. The transmission axis of the first sheet is vertical. The transmission axis of the second sheet is at 30 degrees from vertical. The transmission axis of the third is horizontal. What is the intensity of the light emerging from the third sheet
Answer:
The intensity of light passing from the third polarizer is 3Io/16.
Explanation:
The law of Malus is given by
[tex]I=I_o cos^2\theta[/tex]
Let the incident intensity of light is Io.
The intensity of light passing from the first polarizer is
[tex]I' = \frac{I_o}{2}[/tex]
The intensity of light passing from the second polarizer is
[tex]I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}[/tex]
The intensity of light passing from the third polarizer is
[tex]I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}[/tex]
what is simple definition of democracy
it's a form of government where people elect their representatives
Answer:
The word democracy itself means rule by the people.
what is newtons 2nd law
According to the Newton's second law :- The acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
The velocity of an object traveling in a circle is quadrupled and its radius is tripled The acceleration of this object will change by factor of?
Answer:
The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.
Sample Problem #1
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 900 kg
v = 10.0 m/s
R = 25.0 m
Requested Information:
a = ????
Fnet = ????
To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
Sample Problem #2
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 95.0 kg
R = 12.0 m
Traveled 1/4-th of the circumference in 2.1 s
Requested Information:
v = ????
a = ????
Fnet = ????
To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v = d / t
v = (0.25 • 2 • pi • R) / t
v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (8.97 m/s)2 / (12.0 m)
a = (80.5 m2/s2) / (12.0 m)
a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
PLEASE HELP MEE THIS IS DUE IN 45 MINS
Answer:
The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Explanation:
This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:
[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)
Where:
[tex]m[/tex] - Mass of the car, in kilogram.
[tex]v[/tex] - Initial velocity, in meters per second.
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]s[/tex] - Travelled distance, in meters.
Then we derive an expression for the distance travelled by the vehicle:
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]
[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]
As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
The hottest ordinary star in our galaxy has a surface temperature of 53,000 K. Part A What is the peak wavelength of its thermal radiation
Answer:
[tex]\lambda=5.46\times 10^{-8}\ m[/tex]
Explanation:
The hottest ordinary star in our galaxy has a surface temperature of 53,000 K.
We need to find the peak wavelength of its thermal radiation.
Using Wein's law,
[tex]\lambda T=2.898\times 10^{-3}\\\\\lambda=\dfrac{2.898\times 10^{-3}}{53000}\\\\=5.46\times 10^{-8}\ m[/tex]
So, the peak wavelength of its thermal radiation is equal to [tex]5.46\times 10^{-8}\ m[/tex].
An eagle is flying horizontally at a speed of 3.40 m/s when the fish in her talons wiggles loose and falls into the lake 5.50 m below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the x-direction and that the y-direction is up.)
Answer:
The velocity of the fish when it hits the water is:
v = 10.93 m/s and 71.88 ° below the x-direction.
Explanation:
Let's find the velocity of the fish in the y-direction.
[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]
Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:
[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]
[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]
[tex]v_{fy}=-10.39 \: m/s[/tex]
Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]
[tex]v=10.93 \: m/s[/tex]
And the direction will be:
[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]
[tex]\theta=71.88^{\circ}[/tex]
The angle is 71.88 ° belox the x-direction.
I hope it helps you!
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration
Answer:
i) the minimum acceleration to take off is 22500 km/h²
ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) required force that the engine must exert to attain acceleration is 625 kN
Explanation:
Given the data in the question;
mass of plane m = 360,000 kg
take of speed v = 300 km/hr = 83.33 m/s
i)
What should be the minimum acceleration to take off if the length of the runway is 2.00 km
from Newton's equation of motion;
v² = u² + 2as
we know that a plane starts from rest, so; u = 0
given that distance S = 2 km
we substitute
(300)² = 0² + ( 2 × a × 2 )
90000 = 4 × a
a = 90000 / 4
a = 22500 km/h²
Therefore, the minimum acceleration to take off is 22500 km/h²
ii) At this acceleration, how much time would the plane need from starting to takeoff.
from Newton's equation of motion;
v = u + at
we substitute
300 = 0 + 22500 × t
t = 300 / 22500
t = 0.0133 hrs
Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) What force must the engines exert to attain this acceleration
we know that;
F = ma
acceleration a = 22500 km/hr² = 1.736 m/s²
so we substitute
F = 360,000 kg × 1.736 m/s²
F = 624960 N
F = 625 kN
Therefore, required force that the engine must exert to attain acceleration is 625 kN
Your hand and wrist curl in toward the center of your body (chest and stomach) to prepare to throw the frisbee.
O True
O False
True
Hope this helps! :)
Answer:
true because when trow the frisbee gives u level
Consider the following possibilities and select the correct choice.
1. Tx Ty > Tz
2. Tx Ty < Tz
3. Tx Ty = Tz
Answer:
Tx not but mybe
Explanation:
for that reason its just trying to help
When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
the period of oscillation of the given object is 0.14 s
Explanation:
Given;
mass of the object, m = 3 kg
extension of the spring, x = 0.085 m
The spring constant is calculated as follows;
[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]
The angular speed of a 4 kg object is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]
Therefore, the period of oscillation of the given object is 0.14 s
According to ____________ , the randomness of the universe is constantly increasing.
a. The first law of thermodynamics
b. The zeroth law of thermodynamics
c. The second law of thermodynamics
Answer:
According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?
Explanation:
So answer option C. Have a great summer.
I NEED THE ANSWER QUICK PLEASEE
Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 6.7 X 10 -27 Kg.
Answer:
v₂ /v₁ = 2.3 10⁺²
Explanation:
The energy is conserved so the total potential energy must be transformed into kinetic energy
K = U
½ m v² = q ΔV
v = [tex]\sqrt{\frac{2q \Delta V}{m} }[/tex]
a) Let's find the speed of the electron
m = 9.1 10⁻³¹ kg
as they do not indicate the value of the power difference, we will assume that ΔV = 1 V is worth one
v = [tex]\sqrt{ \frac{2 \ 1.6 \ 10^{-19} \ 1}{9.1 \ 10^{-31}} }[/tex]
v = [tex]\sqrt {0.3516 \ 10^{12}}[/tex]
v1 = 0.593 10⁶ m / s
b) the velocity of a hydrogen ion
M = M_H + m
M = 1.673 10⁻²⁷ + 9.1 10⁻³¹
M = 1.67391 10⁻²⁷ kg
M = 1.67 10⁻²⁷ kg
v = [tex]\sqrt{ \frac { 2 \ 1.6 \ 10^{-19} \ 1}{1.67 \ 10^{-27}} }[/tex]
v = [tex]\sqrt{ 1.916167 \ 10^8 }[/tex]
v₂ = 1.38 10⁴ m / s
the relationship between these speeds is
v₂ / v₁ = 1.38 10⁴ / 0.593 10⁶
v₂ /v₁ = 2.3 10⁺²
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
A Child stands on the bus Remains Still When The bus is at rest. When the bus moves forward AndeaThe bus is at rest. When the bus moves forward And then slows down, the children the Contnues moving forward at the original speed. This is an example of
Answer:
inertia
Explanation:
Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field component transmitted
Answer:
the rms value of the electric field component transmitted is 3.295 V/m
Explanation:
Given;
intensity of the unpolarized light, I = 0.0288 W/m²
For unpolarized light, the relationship between the amplitude electric field and intensity is given as;
[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]
The relationship between the rms value of the electric field and the amplitude electric field is given as;
[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]
Therefore, the rms value of the electric field component transmitted is 3.295 V/m
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible
Answer:
By the Pythagorean Theorem the distances from the speakers os
5 and 5.5 (rounding) meters - let y be the wavelength in the solution
n y = 5 n is number of wavelengths from speaker
(n + m) y = 5.5 m must be integral for constructive interference
m y = .5 subtracting equations
m = 2 and y = ,25 for the above conditions
(n + 2) y = 5.5 substituting for m
n = 5.5 / .25 - 2 = 20
f = v / y using frequency of sound
f = 340 / .25 = 1360 / sec for lowest frequency
Check: D1 = y n = ,25 * 20 = 5
and D2 = .25 * 22 = 5.5 for the distances traveled
all
What is Velocity
Answer:
noun
the speed of something in a given direction.
"the velocities of the emitted particles"
(in general use) speed.
"the tank shot backwards at an incredible velocity"
Similar:
speed
pace
rate
tempo
momentum
impetus
swiftness
swift/fast pace
fastness
quickness
speediness
rapidity
briskness
expeditiousness
expedition
dispatch
acceleration
clip
fair old rate
fair lick
steam
nippiness
fleetness
celerity
ECONOMICS
the rate at which money changes hands within an economy.
noun: velocity of circulation; plural noun: velocities of circulation
Answer:
i hope this helps you
Explanation:
hii
A 105 kg astronaut carrying a 16 kg tool bag finds himself separated from his spaceship by 18 m and moving away from the spaceship at 0.1 m/s. To get back to the spaceship, he throws the tool bag away from the spaceship at 4.5 m/s (relative to the station). How long (in s) will he take to return to the spaceship
Answer:
[tex]T=22.5sec[/tex]
Explanation:
From the question we are told that:
Mass of astronaut [tex]m_a=105kg[/tex]
Mass of tool [tex]m_t=16kg[/tex]
Distance [tex]d=18m[/tex]
Velocity of separation [tex]v_s= 0.1m/s[/tex]
Velocity of tool bag [tex]v_t=4.5m/s[/tex]
Generally the equation for momentum is mathematically given by
[tex]P=mv[/tex]
Therefore
Initial Momentum before drop
[tex]P_1=0.1(105+16)[/tex]
[tex]P_1=12.1[/tex]
Initial Momentum after drop
[tex]P_2=-16(4.5)+105V[/tex]
Therefore
Since [tex]P_1=P_2[/tex]
[tex]-72+105V=12.1[/tex]
[tex]V=0.8m/s[/tex]
Generally the equation for Time T is mathematically given by
[tex]T=\frac{d}{V}[/tex]
[tex]T=\frac{18}{0.8}[/tex]
[tex]T=22.5sec[/tex]
How can a wire become magnetic?
add a resistor
point it north
heat it up
run a current through it
Answer:
Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.
Explanation:
A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?
Answer:
[tex]d =3.7*10^{-3} m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.2kg[/tex]
Spring constant [tex]\mu=22Nm^{-1}[/tex]
Amplitude [tex]A=5cm=0.05m[/tex]
Generally the equation for displacement d is mathematically given by
[tex]d = Asin(\omega t)[/tex]
Where
[tex]\omega=angular\ velocity[/tex]
[tex]\omega=\sqrt{k/m}[/tex]
[tex]\omega=\sqrt{22/1.2}[/tex]
[tex]\omega=4.2817rads^{-1}[/tex]
Therefore
[tex]d = 0.05*sin(4.2817*1)[/tex]
[tex]d =3.7*10^{-3} m[/tex]
What does it mean when work is positive?
O Velocity is greater than kinetic energy
O Kinetic energy is greater than velocity
The environment did work on an object.
O An object did work on the environment.
d. An object did work on the environment.
Explanation:Work is defined in many contexts. Some of these are;
i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.
ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.
In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).
Answer:
its c
Explanation:
The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diameter of 0.88 m. If the car continues to decelerate at this rate, how far does it go
Answer:
-2.869 rad/s2
Explanation:
Data given:
speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)
Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds
And for every 1km, there will be 1000m.
So, speed of 95.0 km/h = 26.389 m/s
speed, vi = r ω (radius X angular velocity)
angular velocity, ωi = v/r
ωi = 26.389 m/s ÷ half of 0.88 m diameter
= 59.975 rad/s
decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)
= 16.667m/s
The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter
= 37.879rad/s
How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ
= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)
= -2.869 rad/s2
find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
Given the following data;
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
[tex] Resistance = P \frac {L}{A} [/tex]
Where;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]
[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]
Resistance = 9.95 Ohms
Which of the following would experience induced magnestism mostly easily
Answer:
Copper would experience induced magnestism mostly easily.
Explanation:
Permalloy would experience induced magnetism most easily.
I don't know the options but that is correct too!!
